Filtering by fold change
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@peter-robinson-529
Last seen 10.2 years ago
Dear List, I guess this is a pretty basic question, but I have not found an answer in the archives. I would like to take a matrix representing m hybridizations and sort it to extract the 2000 genes with the maximal fold change as a preliminary step of a further analysis. Is it possible to use the geneFilter to extract a fixed number of the "top" genes as above rather than genes satisfying some fixed criterion? Thanks Peter
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Anthony Bosco ▴ 500
@anthony-bosco-517
Last seen 10.2 years ago
The easiet way is as follows for Affymetrix data. Load and analyse data data<-ReadAffy() eset<-rma(data) exp<-eset@exprs Calculate fold change (m) values for 2 chips m<-exp[,1]-exp[,1] Ask for top up and down regulated genes X<-names(m[abs(m)>2]) Keep adjusting m cut off (ie 2 in above example) to get desired number of genes Filter data exp.subset<-exp[X,] Alternatively Sort on fold change m.sort <- sort(m,dec=T) top.genes <- m.sort[1:1000] exp.subset<-exp[top.genes,] Cheers Anthony -- ______________________________________________ Anthony Bosco - PhD Student Institute for Child Health Research (Company Limited by Guarantee ACN 009 278 755) Subiaco, Western Australia, 6008 Ph 61 8 9489 , Fax 61 8 9489 7700 email anthonyb@ichr.uwa.edu.au ______________________________________________ [[alternative HTML version deleted]]
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On Mon, 2004-06-28 at 07:47, Anthony Bosco wrote: > The easiet way is as follows for Affymetrix data. > > Load and analyse data > > data<-ReadAffy() > > eset<-rma(data) > > exp<-eset@exprs > Calculate fold change (m) values for 2 chips > > m<-exp[,1]-exp[,1] Err, isn't this equivalent to m <- x - x which always produce 0. Besides I think the definition of Fold Change is the difference of the group _means_. Using exp is bad idea as it is also an R function, see help("exp"). > Ask for top up and down regulated genes > > X<-names(m[abs(m)>2]) Your solution can produce spurious results. You should surround the abs(m) > 2 with which(). See below : > x <- 1:5 > names(x) <- letters[1:5] > x a b c d e 1 2 3 4 5 > > names( x > 2.5 ) [1] "a" "b" "c" "d" "e" > > names( which( x > 2.5 ) ) [1] "c" "d" "e" Note : There is the issue of translating from log2 base if you use RMA. So a FC of 2 would correspond to log2(FC) of 1. > Keep adjusting m cut off (ie 2 in above example) to get desired number of genes > > Filter data > > exp.subset<-exp[X,] > > Alternatively Sort on fold change > > m.sort <- sort(m,dec=T) > top.genes <- m.sort[1:1000] > > exp.subset<-exp[top.genes,] It is much easier and faster to use indices than matching rownames. data <- matrix( rnorm(100000), nc=10 ) rownames(data) <- paste("g", 1:10000, sep="") > w <- which(abs(rowMeans(data)) > 0.5) > names.w <- rownames(data)[w] length(w); length(names.w) [1] 1173 [1] 1173 > system.time( sub1 <- data[ w, ] ) [1] 0 0 0 0 0 > system.time( sub1 <- data[ names.w, ] ) [1] 0.47 0.00 0.48 0.00 0.00
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@kasper-daniel-hansen-459
Last seen 10.2 years ago
On Sun, Jun 27, 2004 at 09:19:33PM +0200, peter robinson wrote: > Dear List, > > I guess this is a pretty basic question, but I have not found an answer in the > archives. I would like to take a matrix representing m hybridizations and > sort it to extract the 2000 genes with the maximal fold change as a > preliminary step of a further analysis. Is it possible to use the geneFilter > to extract a fixed number of the "top" genes as above rather than genes > satisfying some fixed criterion? No. Genefilter operates on a gene-by-gene basis (at least when I last looked at the function). There are (of course) other ways of doing it. -- Kasper Daniel Hansen, Research Assistant Department of Biostatistics, University of Copenhagen
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